I would do the following: First, you want x such that Ps(x)=Pu(x)+0.2*Pu(x). So, define an anonymous function: myf = @(x)(((Psmax./(1+((Psmax/Po)-1)*exp(-ks*x))))-1.2*(Pumax*exp(-ku*x) + Pumin));
This essentially defines the function myf(x)=Ps(x)-(Pu(x)+0.2*Pu(x))
You are then after an x in a way that myf(x) is 0. You can find such x using many methods, such as fzero:
fzero(myf,35)
35 is just a guess for the solution and the function returns 39.6068.
Indeed if you do
a=39.6068;Psmax./(1+((Psmax/Po)-1)*exp(-ks*a))-1.2*(Pumax*exp(-ku*a) + Pumin)
you will get close to zero.
You can see the corresponding values by:
a=39.6068;disp([(Psmax./(1+((Psmax/Po)-1)*exp(-ks*a))) Pumax*exp(-ku*a) + Pumin]);
Does that work?
